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(D) Let the position of girl at $A$ be $(0, 0)$ and girl at $B$ be $(b, 0)$.The girl at $B$ moves with velocity $\vec{V_1} = V_1 \hat{j}$. Her positio

Vedclass · Accepted Answer

$\frac{b}{\sqrt{V_2^2 - V_1^2}}$

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(D) Let the position of girl at $A$ be $(0, 0)$ and girl at $B$ be $(b, 0)$.The girl at $B$ moves with velocity $\vec{V_1} = V_1 \hat{j}$. Her position at time $t$ is $\vec{r_B}(t) = b \hat{i} + V_1 t \hat{j}$.The girl at $A$ moves with velocity $\vec{V_2}$ such that $|\vec{V_2}| = V_2$. Let $\vec{V_2} = V_2 \cos 	heta \hat{i} + V_2 \sin 	heta \hat{j}$.Her position at time $t$ is $\vec{r_A}(t) = (V_2 \cos 	heta) t \hat{i} + (V_2 \sin 	heta) t \hat{j}$.For them to meet,$\vec{r_A}(t) = \vec{r_B}(t)$,so $V_2 \cos 	heta t = b$ and $V_2 \sin 	heta t = V_1 t$.From the second equation,$\sin 	heta = V_1 / V_2$. Thus,$\cos 	heta = \sqrt{1 - (V_1/V_2)^2} = \frac{\sqrt{V_2^2 - V_1^2}}{V_2}$.Substituting $\cos 	heta$ into the first equation: $V_2 \left( \frac{\sqrt{V_2^2 - V_1^2}}{V_2} ight) t = b$.Therefore,$t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$.

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